Question

Writing an equation of an eclipse given the foci and major axis length

Answer 1

Answer:

The equation of the ellipse is:

$\begin{equation*} \frac{(x-3)^2}{64}+\frac{(y+3)^2}{55}=1 \end{equation*}$

Step-by-step explanation:

Remember that the general equation of an ellipse is:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

Where:

• (h,k), are the coordinates of the center

,

• a, is the major axis lenght

,

• b, is the minor axis lenght

We also have that the equation for the foci is:

$F(h\pm c,k)$

And we also have that:

$c^2=a^2-b^2$

Since we have two foci, we can find the values of h and c as following:

$\begin{gathered} F(h\pm c,k) \\ \\ (6,-3)\rightarrow h+c=6 \\ (0,-3)\rightarrow h-c=0 \end{gathered}$

We'll have the following system of equations:

$\begin{cases}h+c=6 \\ h-c={\text{ }0}\end{cases}$

Adding up both equations and solving for h,

$\begin{cases}h+c=6 \\ h-c={\text{ }0}\end{cases}\rightarrow2h=6\rightarrow h=3$

Using this h value in the first equation and solving for c,

$h+c=6\rightarrow3+c=6\rightarrow c=3$

We'll have that the solution to this particular system of equations is:

$\begin{gathered} h=3 \\ c=3 \end{gathered}$

Now we know the value of c we can calculate the value of b as following:

$\begin{gathered} c^2=a^2-b^2 \\ \rightarrow b^=\sqrt{a^2-c^2} \\ \rightarrow b=\sqrt{8^2-3^2} \\ \\ \Rightarrow b=\sqrt{55} \end{gathered}$

With all these calculations, we'll have all the values we need to put together the equation:

$\begin{gathered} h=3 \\ k=-3 \\ a=8 \\ b=\sqrt{55} \end{gathered}$

This way, we'll have that the equation of the ellipse is:

$\begin{gathered} \frac{(x-3)^2}{8^2}+\frac{(y-(-3))^2}{(\sqrt{55})^2}=1 \\ \\ \Rightarrow\frac{(x-3)^2}{64}+\frac{(y+3)^2}{55}=1 \end{gathered}$